How far up the ladder can you climb before it begins to slip?

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The discussion revolves around calculating how far up a 9 m ladder, leaning against a frictionless wall at a 47° angle, a person weighing 74 kg can climb before it slips. The ladder itself weighs 33 kg, and the coefficient of static friction between the ladder and the ground is 0.41. The equilibrium condition is established, leading to the equation for torque, where the forces from the wall and friction are balanced. The correct maximum distance calculated for climbing is approximately 3.715 m up the ladder. Clarifications are made regarding the nature of static friction, emphasizing that it has a maximum value and can lead to slipping if exceeded.
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A uniform ladder of length 9 m leans against a frictionless vertical wall making an angle of 47° with the ground. The coefficient of static friction between the ladder and the ground is 0.41.

If your mass is 74 kg and the ladder's mass is 33 kg, how far up the ladder can you climb before it begins to slip?


Given:
L = 9 m
M = 74 kg
m = 33 kg
μ = 0.41
θ = 47°
x = unknown


Noted:
FWL = Force from Wall to Ladder
Ff = Force of Friction
N = Normal ForceI think I've solved it, but I just wanted to be sure. I was trying to get some practice in for my test tomorrow and found this question on these forums unanswered.

Since nothing is moving, the system is in Equilibrium. So net torque = 0, or counter clockwise torques = clockwise torques. Equation for torque; t = Force x Moment Arm

Mg(xL cosθ) + mg(0.5L cosθ) = FWL(L sinθ)
The L cancels out from both sides, and the g cosθ is factored out of left side.

g(xM + 0.5m) cosθ = FWL sinθ
Divide both sides by sinθ

g(xM + 0.5m) / tanθ = FWL
Since system is at equilibrium FWL = Ff = μN = μ(M + m)g

g(xM + 0.5m) / tanθ = μ(M + m)g
Multiple both sides by tanθ, and g cancels from both sides

xM + 0.5m = μ(M + m) tanθ
xM = μ(M + m) tanθ - 0.5m
x = [μ(M + m) tanθ - 0.5m] / M = 0.4127...

Max length up ladder = xL ≈ 3.715 m
 
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liquidheineken said:
Since system is at equilibrium FWL = Ff = μN = μ(M + m)g


Your solution is correct but the sentence above is not.

It should be

Since system is at equilibrium FWL = Ff ≤ μN = μ(M + m)g
 
can you elaborate on that? I was simply plugging in the equation for Friction at that point. Why is Ff ≤ μN?
 
Matter of careful wording: The system is also at equilibrium when xL = 0 but then ##F_f \ne \mu N##
 
liquidheineken said:
can you elaborate on that? I was simply plugging in the equation for Friction at that point. Why is Ff ≤ μN?
The force of static friction is a force that prevents sliding. It does not have a definite value, only a maximum: depending on the surfaces in contact, the maximum value is μsN, where μs is the coefficient of static friction. If the other forces exceed the static friction, the object starts to slide and is not in equilibrium any more. The friction becomes kinetic, and equal to μkN
 
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